how many moles of your excess reactant remain at the end of the reaction?
if you start with 54g of lithium iodide and 5.0 g of oxygen, how many moles of reactant remain at your solution
hcbiochem
Balanced equation:
4 LiI + O2 --> 2 Li2O + 2 I2
Moles LiI = 54 g / 133.85 g/mol = 0.403 mol LiI
Moles O2 = 5.0 g / 32.0 g/mol = 0.156 mol O2
0.403 mol LiI X (1 mol O2/ 4 mol LiI) = 0.101 mol O2
Because you begin with more O2 than this, LiI is the limiting reactant, and 0.156 - 0.101 = 0.055 mol of O2 remains when the reaction is complete.